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Unit vector is given by calculating the cross product of a=2i+j+k and b=3i-j+2k,that is a×b = {(1×2) - (1×-1)}i - {(2×2)-(1×3)}j + {(2×-1)-(1×3)} j giving 3i-j-5k
Unit vector = magnitude of |3i-j-5k| =sqrt (35)
Unit vector = {(1/sqrt35)3i} - {(1/sqrt35)j} - {(1/sqrt35)5k}
The angle between a and b, is given by first getting the magnitude of a, b and AB, then using cosine rule. Magnitude of |a|=|2i+j+k| = sqrt 6, |b|=|3i-j+2k|= sqrt14 and |AB|= |i-2j+k|=sqrt 6
Cosine rule =a^2 =b^2 + c^2 - 2bccosA
6=6+14-18.33cosA giving A =40.2°
YitzhakRabin answered the question on October 24, 2018 at 23:41
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